a.
When I solved for this problem, I first used the ${s_{n+1}} – {s_n} = {a_{n+1}}$ to solve for a number of nth terms in ${a_n}$. Using the series in ${a_n}$, I calculated that the closed formula for the series when n > 1 was $\frac{-109}{4n^2 + 40n + 99}$.
b.
For this part, I took the limit of ${a_n}$ to find the value, but first I took the constant out, giving me $\frac{1}{4n^2 + 40n + 99}$. Substituting in the infinity, I got $\frac{1}{\infty} = 0$.
Both these answers were considered to be incorrect, but I'm not sure where I went wrong. Any kind of guidance would be greatly appreciated.
Best Answer
For part (a), you are following the correct procedure, although you have found $a_{n+1}$, not $a_n$ as asked for. Therefore, \begin{align} a_n&=s_n-s_{n-1} \\ &=\frac{14-9n}{2n+9}-\frac{14-9(n-1)}{2(n-1)+9} \\ &=\frac{14-9n}{2n+9}- \frac{23-9n}{2n+7} \\ &=-\frac{109}{(2n+9)(2n+7)}. \end{align}
Alternatively, you can replace $n$ with $n-1$ in your answer.
For part (b), by definition, $$\sum_{n=1}^\infty a_n = \lim_{n \to \infty}s_n=\lim_{n \to \infty}\frac{14-9n}{2n+9}=-\frac{9}{2}.$$
Hope this helps!